I'm getting familiar with basic indefinite integrals and these are the hardest ones I've met so far:
$\int \frac{\sqrt{x}}{x^2(\sqrt{x+1}+\sqrt{x})}dx$
$\int \frac{\sqrt[3]{x+2}-\sqrt[3]{x}}{x^2(\sqrt[3]{x+2}+\sqrt[3]{x})}$
Any hints? Please note that the course I am taking does not anticipate usage of hyperbolic functions. I am not familiar with them.
The first integral I attempted:
$\int \frac{\sqrt{x}}{x^2(\sqrt{x+1}+\sqrt{x})}dx = \int \frac{\sqrt{x}(\sqrt{x+1}-\sqrt{x})}{x^2(\sqrt{x+1}+\sqrt{x})(\sqrt{x+1}-\sqrt{x})}dx = \int \frac{\sqrt{x}\sqrt{x+1}-x}{x^2}dx$
Now I can split it into two integrals. Problem is with:
$\int \frac{\sqrt{x}\sqrt{x+1}}{x^2}dx$
and the major problem is that I don't know how to solve integrals that have some distinct roots with different values inside those roots. Second task's integral seems even harder.
If speaking of "different values under roots", I am only familiar with how to solve such integrals:
$\int \frac{(\sqrt{\frac{x+2}{x-1}}-1)^2}{3(\sqrt{\frac{x+2}{x-1}}+2)}dx$
because there's simple algorithm that I can follow to solve it.
Hints, tips, advices appreciated. Thanks.
EDIT: $\int \frac{\sqrt{x}\sqrt{x+1}}{x^2}dx = \int \frac{\sqrt{x^2+x}}{x^2}dx = \int \frac{x^2 + x }{x^2\sqrt{x^2+x}}dx = \int \frac{x+1}{x\sqrt{x^2+x}}dx$
And now with Euler's substitution should work?
Hint: For the integral $$\int \frac{\sqrt{x^2+x}}{x^2}dx$$ substitute $$\sqrt{x^2+x}=x+t$$ it is the Eulerian substitution. Then we get by squaring $$x=\frac{t^2}{1-2t}$$ and $$dx=-2\,{\frac {t \left( -1+t \right) }{ \left( -1+2\,t \right) ^{2}}}dt$$